Illustrating the Oxidation States of Mn & V
Illustrating the Oxidation States of Mn & V
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Procedure © KCl http://hk.geocities.com/fatherofchemistry
Discussion
Part A: Making Mn(VI) from Mn(VII) and Mn(IV) 1.> Explain why only one of the three mixtures reacted to give green Mn(VI).
[ANS] By Le Chatelier's Principle, only the alkaline medium will shift the equilibrium to right and yield green MnO42-.
2.> What happened when acid was added to Mn(VI)? Explain.
[ANS] The solution changes from green to purple again. The acid removes OH- on the left side and the decrease in [OH-] favours the backward reaction to form MnO4- (purple).
Part B: Making Mn(III) from Mn(II) and Mn(VII) 1.> Explain what happened when the Mn(III) solution is diluted.
[ANS] When Mn3+ is diluted, i.e. more water is added on right side, the equilibrium will shift leftward, purple MnO4- will form.
Part C: Making Mn(III) from Mn(II) and Mn(IV) 1.> What is different about the conditions of this experiment (part C) compared with last (part B) which makes its success less likely?
[ANS] The condition of part C is different from part B in which it's in an alkaline medium in contrary to acidic medium in part B. The OH- will combine with Mn3+ to form insoluble Mn(OH)3 which unfavours this reaction.
Part D: Vanadium
Ion (hydrated) VO3- / VO2+ VO2+ V3+ V2+
Colour yellow blue green violet
Oxidation State +5 +4 +3 +2 Test Observations Summary of reaction
NH4VO3 + acid white solid dissolved to a yellow solution
V(V) + Zn solution changes from yellow via green to blue, then to green and finally deep blue, gas evolves with a rotten-egg odour
V(II) + manganate(VII) solution changes from violet to green, then to blue and finally yellow
1.> V(V) + sulphite
2.> Add V(II) solution changes from yellow via green to blue
blue solution turns green
V(V) +
© KCl http://hk.geocities.com/fatherofchemistry
Procedure © KCl http://hk.geocities.com/fatherofchemistry
Discussion
Part A: Making Mn(VI) from Mn(VII) and Mn(IV) 1.> Explain why only one of the three mixtures reacted to give green Mn(VI).
[ANS] By Le Chatelier's Principle, only the alkaline medium will shift the equilibrium to right and yield green MnO42-.
2.> What happened when acid was added to Mn(VI)? Explain.
[ANS] The solution changes from green to purple again. The acid removes OH- on the left side and the decrease in [OH-] favours the backward reaction to form MnO4- (purple).
Part B: Making Mn(III) from Mn(II) and Mn(VII) 1.> Explain what happened when the Mn(III) solution is diluted.
[ANS] When Mn3+ is diluted, i.e. more water is added on right side, the equilibrium will shift leftward, purple MnO4- will form.
Part C: Making Mn(III) from Mn(II) and Mn(IV) 1.> What is different about the conditions of this experiment (part C) compared with last (part B) which makes its success less likely?
[ANS] The condition of part C is different from part B in which it's in an alkaline medium in contrary to acidic medium in part B. The OH- will combine with Mn3+ to form insoluble Mn(OH)3 which unfavours this reaction.
Part D: Vanadium
Ion (hydrated) VO3- / VO2+ VO2+ V3+ V2+
Colour yellow blue green violet
Oxidation State +5 +4 +3 +2 Test Observations Summary of reaction
NH4VO3 + acid white solid dissolved to a yellow solution
V(V) + Zn solution changes from yellow via green to blue, then to green and finally deep blue, gas evolves with a rotten-egg odour
V(II) + manganate(VII) solution changes from violet to green, then to blue and finally yellow
1.> V(V) + sulphite
2.> Add V(II) solution changes from yellow via green to blue
blue solution turns green
V(V) +